$$z_k=\cos\left( \frac{2k\pi}{10}\right)+i\sin\left(\frac{2k\pi}{10}\right);k\in \{1,2,3,\ldots,9\}$$ then find the value of:$$\frac{|1-z_1||1-z_2||1-z_3|\ldots|1-z_9|}{10}$$
Answer: 1
My attempt:
$$\text{let, }\frac{|1-z_1||1-z_2||1-z_3|\ldots|1-z_9|}{10}=p$$ $$p^2=\frac{|1-z_1|^2|1-z_2|^2|1-z_3|^2\ldots|1-z_9|^2}{100}$$ $$\implies p^2=\frac{(1-z_1)(1-\overline{z_1})(1-z_2)(1-\overline{z_2})\ldots(1-z_9)(1-\overline{z_9})}{100}\text{, since } z\overline{z}=|z|^2$$ $$=\frac{\{(1-z_1)(1-z_2)\ldots (1-z_9)\}\{(1-\overline{z_1})(1-\overline{z_2})\ldots (1-\overline{z_9})\}}{100}$$
since $z_k$ is the $10^{th}$ root of unity hence $x^{10}-1=(x-1)(x-z_1)\ldots(x-z_9)$ $$\implies \frac{x^{10}-1}{x-1}=(x-z_1)(x-z_2)\ldots(x-z_9)$$ $$x=1, \frac{1^{10}-1}{1-1}=(1-z_1)(1-z_2)\ldots(1-z_9)=0/0$$
so you see at this point i am stuck, how do i solve this further
Very much the right idea, but the execution was too complicated. Note that $$x^9+x^8+\cdots +x+1=(x-z_1)(x-z_2)\dots (x-z_9),$$ and let $x=1$.